Math Is Fun Forum

Why not? The beauty of math is that 'all roads lead to Rome' I mean the solution of a problem (Rome) could be reached by following different logical paths (well-done roads).

Of course, you asked a correct question.
We just liked to prove it in its general form:
## tan(70) = 2tan(50) + tan(20)
## tan(a) = 2tan(b) + tan(c)
The two sides are equal if'
a+c=90 and b=a-c

Similarly:
## 2tan(70) = tan(80) - tan(10)
## 2tan(b) = tan(a) - tan(c)
The two sides are equal if'
a+c=90 and b=a-c

The trick is that:
1 - tan(70)*tan(20) = 0
1 - tan(80)*tan(10) = 0

1 - tan(a)*tan(c) = 0 , if a+c = 90 , see the denominator of the identity below:

tan(a+c) = tan(90) = [ tan(a) + tan(c) ] / [ 1 - tan(a)*tan(c) ] = ∞  {a fraction equals infinity, if its denominator=0 and its denominator≠0}

Thank you.

It is like... tan(65) = 2tan(45) + tan(25) [wrong] Thanks to Phrzby Phil

It is like... tan(65) = 2tan(40) + tan(25)

I just wonder how an unbreakable encryption can be made breakable at certain sides (the right destinations), and it is supposed to be always unbreakable otherwise, despite the presence of all sorts of spies in all sides!

I thought that scalar could be an algebraic number that is positive or negative. It seems it has to be, by definition, a positive number only (including 0).
In this respect, I am afraid that my post #4 is wrong. What I called scalar on the X, Y or Z axis is also a vector but in a 1D space. Sorry for this mistake.

A vector can be seen, for example, as 2 scalars (in a 2D space, as a scalar on the X axis and another on the Y axis) or 3 scalars (in 3D space, on X, Y and Z) which defined it.
In 2D space, adding two vectors is simply adding their 2 scalars on X also their two scalers on Y. These two sums define the resultant vector.

+1

I liked to prove numerically that your graphical solution is right and your AE+BG=5.2915 is indeed the smallest sum to get an equilateral triangle.

By using the solver of Excel, I got:
AE+BG= 5.291502622
Area = 1.484614945

Please note that on my post #14, equ_3 is wrong. Also, by shifting the origin A to C, the equations become simpler.

But I am not sure if showing here what I did could interest you or anyone else. The last equation happened to be non-linear, so I had to solve it by Excel’s solver while finding the minimum sum by trial and error (thanks to Excel).

Kerim

Note:
I couldn't see your two images:
post #13, https://i.imgur.com/bs1yfhul.jpg
post #17, https://i.imgur.com/4BS9AQTl.jpg

I don't think our dear guest, Johntom, can solve his interesting exercise, analytically.

Here is what I did:
A(0.0)
C(4,0)
B(6,0)
E(m,n)
G(p,q)

Therefore:
AE = sqrt(m^2 + n^2)
BG = sqrt[(6-p)^2 + q^2]

EC^2 = (4-m)^2 + n^2
CG^2 = (p-4)^2 + q^2
EG^2 = (p-m)^2 + (q-n)^2

equ_1: EG^2 = EC^2
           (p-m)^ + (q-n)^2 = (4-m)^2 + n^2
           p^2 - 2pm + m^2 + q^2 - 2qn + n^2 = 16 - 8m + m^2 + n^2
res_1:  p^2 - 2pm + q^2 - 2qn = 16 -8m

equ_2: EG^2 = CG^2
           (p-m)^ + (q-n)^2 = (p-4)^2 + q^2
           p^2 - 2pm + m^2 + q^2 - 2qn + n^2 = p^2 - 8p + 16 + q^2
           - 2pm + m^2 - 2qn + n^2 = - 8p + 16
res_2: 2pm - m^2 + 2qn - n^2 = 8p - 16

From res_1 and res_2
equ_3: p^2 + q^2  = 8p - 8m
res_3: m =  p -  (p^2 + q^2) / 8

From res_1 and equ_3:
          p^2 + q^2 = 16 -8m + 2pm + 2qn
          8p - 8m = 16 -8m + 2pm + 2qn
          8p = 16 + 2pm + 2qn
          2pm = 8p - 16 - 2qn
          pm = 4p - 8 - qn
res_4: m = (4p - 8 - qn) / p

From res_3 and res_4:
          p - (p^2 + q^2) / 8 = (4p - 8 - qn) / p
          p - (p^2 + q^2) / 8 = 4 - 8/p - qn/p
          qn/p = - p + (p^2 + q^2) / 8 + 4 - 8/p
          qn = - p^2 + p*(p^2 + q^2) / 8 + 4p - 8
res_5: n = [ - p^2 + p*(p^2 + q^2) / 8 + 4p - 8 ] / q

So far, if I didn't make mistakes, we have m and n in function of p and q.
So, we need another equation which is actually the trick of this exercise:

Let us add F on EG, so that ACF is a right angle. We have now the angles ECF + FCG = pi/3 (of the equilateral triangle ECG)
FCG = pi/3 - ECF
tan(ECF) = (4 -m)/n
tan(FCG) = (p - 4)/q

We know: tan(A-B) = [tan(A)-tan(B)] / [1+tan(A)*tan(B)]
tan(FCG) = tan(pi/3 - ECF) = [ tan(pi/3) - tan(ECF) ] / [ 1 + tan(pi/3)*tan(ECF) ]
tan(FCG) = [ sqrt(3) - tan(ECF) ] / [ 1 + sqrt(3)*tan(ECF) ]
Therefore:
equ_6: (p - 4)/q = [ sqrt(3) - (4-m)/n ] / [ 1 + sqrt(3)*(4-m)/n ]

If we replace m and n from res_3 and res_5 in equ_6, we get an equation with two unknowns only, p and q.
In other words, we can get p=f(q) or q=f(p)

By replacing p, m and n (which are known in function of q) in the function SUM(q)
SUM(q) = AE + GB  = sqrt(m^2 + n^2) + sqrt[(6-p)^2 + q^2]
SUM(q) is minimum when the derivative SUM'(q) = 0

Et Voila.

It is always very good to have on hands as many formulas as possible.

I used to think that each of them was found in order to be used in certain specific applications, in the first place.

But I seldom hear someone mentioning even one application for which a rather complex formula could ease the calculus related to it.

Nothing is nothing
It is like I was nothing before I was forced to exist by a certain Will.
I said 'I was forced' because I had no will while I was nothing